Question

In the equation $\int \frac{dx}{\sqrt{2ax-x^2}}=a^n{\sin }^{-1}(\frac{x}{a}-1).$
Find the value of n.

Answer 1

$I=\int \frac{1}{\sqrt{2ax-x^2}}dx=\int \frac{1}{a^2-{(x-a)}^2}dx$
Put $x-a=u\Rightarrow dx=du$
$\therefore I=\int \frac{1}{\sqrt{a^2-u^2}}du={\sin }^{-1}\frac{u}{a}+c$
$={\sin }^{-1}\frac{(x-a)}{a}+c$
$={\sin }^{-1}(\frac{x}{a}-1)+c$
$\therefore n=0$